Answer
$3$
Work Step by Step
Here, $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{\sqrt {9x+1}}{\sqrt {x+1}}$
This implies that
$\lim\limits_{x \to \infty} \dfrac{\sqrt {9x+1}}{\sqrt {x+1}}=\sqrt{\lim\limits_{x \to \infty} \dfrac{9x+1}{x+1}}$
or, $ \sqrt{\lim\limits_{x \to \infty} \dfrac{9x+1}{x+1}(\dfrac{1/x}{1/x})}=\sqrt{\lim\limits_{x \to \infty} \dfrac{9+1/x}{1+1/x}}$
Thus,
$\sqrt {\dfrac{9+0}{1+0}}=3$
